Chapter 8.1, Problem 8.43P

Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs and what that motion is when the coefficient of static friction between all surfaces is (a) μ_s = 0.40, (b) μ_s = 0.50.

Fig. P8.43

To determine

Find the magnitude of horizontal force P applied at C for which the motion occurs.

Answer to Problem 8.43P

The magnitude of the horizontal force P is $\underset{\_}{62.8N}$.

The system $\underset{\_}{\text{slides}}$.

Explanation of Solution

Given information:

The mass of the blocks A and B is $m=8\text{kg}$.

The coefficient of static friction between the surfaces is ${\mu }_s=0.40$.

Calculation:

Consider the blocks will slide to the right.

Find the weight (W) of the blocks A and B using the relation.

$W=mg$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^2$.

Substitute 8 kg for m and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}W=8\times 9.81\\ =78.48\text{N}\end{array}$

Find the friction force at block A and B as follows.

$\begin{array}{l}{F}_A={\mu }_s{N}_A\\ F_B={\mu }_s{N}_B\end{array}$

Here, the normal force at block A is $N_A$ and the normal force at block B is $N_B$.

Show the free-body diagram of the shelves as in Figure 1.

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_A+N_B-78.48-78.48=0\\ N_A+N_B=156.96\text{N}\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ P-F_A-F_B=0\\ P-{\mu }_s{N}_A-{\mu }_s{N}_B=0\\ P={\mu }_s\left(N_A+N_B\right)\end{array}$

Substitute 0.40 for ${\mu }_s$ and 156.96 N for $\left(N_A+N_B\right)$.

$\begin{array}{c}P=0.40\times 156.96\\ =62.8N\end{array}$

Therefore, the magnitude of the horizontal force P is $\underset{\_}{62.8N}$.

Take moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ P\left(100\right)+N_A\left(93.26\right)-78.48\left(93.26\right)+F_A\left(200\right)=0\\ P\left(100\right)+N_A\left(93.26\right)-7,319.0448+{\mu }_s{N}_A\left(200\right)=0\end{array}$

Substitute 62.8 N for P and 0.40 for ${\mu }_s$.

$\begin{array}{l}62.8\left(100\right)+N_A\left(93.26\right)-7,319.0448+0.40\times N_A\left(200\right)=0\\ 6,280+93.26N_A-7,319.0448+80N_A=0\\ 173.26N_A=1,039.0448\\ N_A=6\text{N}>0\end{array}$

Therefore, the system $\underset{\_}{\text{slides}}$.

To determine

Find the magnitude of horizontal force P applied at C for which the motion occurs.

Answer to Problem 8.43P

The $\underset{\_}{\text{system}\text{rotates}\text{about}\text{point}B}$.

The magnitude of the horizontal force P is $\underset{\_}{73.2\text{N}}$.

Explanation of Solution

Given information:

The mass of the blocks A and B is $m=8\text{kg}$.

The coefficient of static friction between the surfaces is ${\mu }_s=0.50$.

Calculation:

Consider the blocks will slide to the right.

Find the weight (W) of the blocks A and B using the relation.

$W=mg$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^2$.

Substitute 8 kg for m and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}W=8\times 9.81\\ =78.48\text{N}\end{array}$

Find the friction force at block A and B as follows.

$\begin{array}{l}{F}_A={\mu }_s{N}_A\\ F_B={\mu }_s{N}_B\end{array}$

Here, the normal force at block A is $N_A$ and the normal force at block B is $N_B$.

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_A+N_B-78.48-78.48=0\\ N_A+N_B=156.96\text{N}\\ N_B=156.96-N_A\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ P-F_A-F_B=0\\ P-{\mu }_s{N}_A-{\mu }_s{N}_B=0\\ P={\mu }_s\left(N_A+N_B\right)\end{array}$

Substitute 0.50 for ${\mu }_s$ and 156.96 N for $\left(N_A+N_B\right)$.

$\begin{array}{c}P=0.50\times 156.96\\ =78.48N\end{array}$

Therefore, the magnitude of the horizontal force P is $\underset{\_}{78.48N}$.

Take moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ P\left(100\right)+N_A\left(93.26\right)-78.48\left(93.26\right)+F_A\left(200\right)=0\\ P\left(100\right)+N_A\left(93.26\right)-7,319.0448+{\mu }_s{N}_A\left(200\right)=0\end{array}$

Substitute 78.48 N for P and 0.50 for ${\mu }_s$.

$\begin{array}{l}78.48\left(100\right)+N_A\left(93.26\right)-7,319.0448+0.50N_A\left(200\right)=0\\ 7,848+93.26N_A-7,319.0448+100N_A=0\\ 193.26N_A=-528.9552\\ N_A=-2.73\text{N}<0\end{array}$

Therefore, the $\underset{\_}{\text{system}\text{rotates}\text{about}\text{point}B}$.

The value of $N_A=0$.

Take moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ P\left(100\right)+N_A\left(93.26\right)-78.48\left(93.26\right)+F_A\left(200\right)=0\\ P\left(100\right)+N_A\left(93.26\right)-7,319.0448+{\mu }_s{N}_A\left(200\right)=0\end{array}$

Substitute 0 for $N_A$ and 0.50 for ${\mu }_s$.

$\begin{array}{l}P\left(100\right)+\left(0\right)\left(93.26\right)-7,319.0448+0.50\left(0\right)\left(200\right)=0\\ 100P+0-7,316.0448+0=0\\ P=73.2\text{N}\end{array}$

Therefore, the magnitude of the horizontal force P is $\underset{\_}{73.2\text{N}}$.